Assume to the contrary that for every \(\epsilon \in (0, 1)\) , we have \(\rho _{\varphi ,w}(x) \le 1 - \epsilon \)  but \(\Vert x\Vert  = 1\) . Consider an increasing sequence of simple functions \((x_n) \subset B_{(\Lambda _{\varphi ,w})_a}\)  such that \(m(\mathop {\rm supp}{x_n}) < \infty \)  and \(|x_n|\uparrow |x|\)  a.e. Then by [1]} and by theorem on the order-continuous subspace being a norm-closed order-ideal, we see that \(\Vert x_n\Vert _{\varphi ,w} \uparrow \Vert x\Vert _{\varphi ,w} = 1\) . Without loss of generality, we may assume that \(\Vert x_n\Vert _{\varphi ,w} \ge \frac{1}{2}\)  for every \(n \in \mathbb {N}\)  so that \(1 \ge \frac{1}{\Vert x_n\Vert _{\varphi ,w}} - 1 = a_n \downarrow 0\)  as \(n \rightarrow \infty \) .
Now, let \(K = \sup _n\lbrace \rho _{\varphi ,w}(2x_n)\rbrace \) . From the fact that \(x \in (\Lambda _{\varphi ,w})_a\) , we have \(\rho _{\varphi ,w}(kx) <\infty \)  for every \(k>0\)  [2]}. Hence by the monotonicity of \(\rho _{\varphi ,w}\) , we have \(\rho _{\varphi ,w}(2 x_n) \le \rho _{\varphi ,w}(2 x) < \infty \) , which in turn implies that \(K < \infty \) . However, we also have
\(1 = \rho _{\varphi ,w}\left(\frac{x_n}{\Vert x_n\Vert _{\varphi ,w}}\right) = \rho _{\varphi ,w}(2 a_n x_n + (1 - a_n)x_n) &\le & a_n\rho _{\varphi ,w}(2x_n) + (1 - a_n)\rho _{\varphi ,w}(x_n)\\&\le & K a_n + (1- \epsilon )\rightarrow 1 - \epsilon ,\) 
which leads to a contradiction. Therefore, there exists \(\delta \in (0, 1)\)  such that \(\Vert x\Vert _{\varphi ,w} \le 1 - \delta \) .
